# Trig Identies

1. Jun 21, 2010

### wil3

Hello. I am studying second-order differential equations, and I am currently studying simple harmonic oscillations, like those in a spring-block system. When solving the differential equation, I get answers in the form of y^2=

$$x(t)^2=(c_1^2+c_2^2)(cos^2(\omega t)-(c_2^2/(c_1^2+c_2^2))sin^2(\pi/\omega))$$

How can I convert this to the general form of the SHM equation:

$$x(t)^2=(c_1^2+c_2^2)cos^2(\omega t+\phi)$$

where $$cos(\phi)=c_1/\sqrt{c_1^2+c_2^2}$$ and $$sin(\phi)=-c_2/\sqrt{c_1^2+c_2^2}$$

I'm really close here, but I do not know how to use trig identities to convert the squared constant plus squared cosine function into just a single cosine function containing a shift. Thank you.

(PS- I am also trying to use Latex, so check back to see if I've edited the post to make it more readable. Hopefully, you can at least see what I'm going for)

Last edited: Jun 21, 2010
2. Jun 21, 2010

### Gerenuk

In general you can use
$$\cos(x)=\Re e^{ix}$$
and
$$\cos^2 x=\Re e^{ix}\cdot\frac{e^{ix}+e^{-ix}}{2}$$
You get the idea?

Hmm, in your equation the $/\omega$ looks funny. And are you sure that you don't have too many squares in there?

Last edited: Jun 21, 2010
3. Jun 21, 2010

### wil3

No, that's the correct equation. This is how far I got solving back from:

$$x(t)=c_1cos(\omega t)+c_2sin(\omega t)$$

I already used Euler's formula on the solution to the differential equation to get the above equation, which I then manipulated using trig identities to get the equation I printed above. How would I use the formula again to make progress on the problem? Basically, I am looking for a way to show with trig identities that a cosine squared function plus a constant is the same as a cosine squared function with a shifted argument.

4. Jun 21, 2010

### Gerenuk

Then start from the latter equation you quote and use the idea I propose. It's pretty easy then.

5. Jun 21, 2010

### uart

wil3, if your equation in post #3 is correct then your other equation (post #1) is definitely incorrect.